Electrical engineering smart grid, Lecture notes of Electrical Engineering

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ARC LEnth

7 x , | 1l-V24+a ve n two ways. Usually it pays to consider substitution as the first option and here obviously the root causes t ered expression and check whether we can see this derivative next to dy. That is not true here, but the mor y=Vv2+a 2 7 2=y-2 2 - 39,3 x y° —2 2y” — dy ——— dr = dx = 2ydy - | toy dy = [ —— dy. = r=2Hy=2 2 l-y 2 l-y r=TrRy=3 f 20 dr 3 (2+ 4)(2-4)(2-1) ‘d type belonging to a box, so we try to think of some substitution. We do not like those reciprocal x, so acceed, because the derivative of 1/x is (up to the sign) equal to 1/x?, which is exactly what is available 2n substitute the limits. —i1 Y= dy = = dx -| ~20 dy (2+ y)(2 — y)(2y — 1) / 20 dx 7 P(2+2)2-2)(F-1) -/ 20 dy (y + 2)(y — 2)(2y — 1)" | fractions. Because we have distinct linear factors, the unknown constants can be easily obtained using : the two from the last factor. The linear factor are then easily integrated using substitution, for instance / 20dxr [2 +a dy -¥# f dy w(2+2)(2-2)(2-1) J yt2 > Sf y-2 FF 2-1 [ovis ae. ive to try some substitution and hope for the best. There are two reasonable pos: y= 1 — et [> 1 =e dir = | dy = —2e¥ de =| vig y y dz = ge dy = 555 sa oD ft, so we have to recall the box "integrals with roots" and take the square root of z= VV [vizmae~ fHay-| 2 dy = 2zdz 22? =/a <22dz = Po In Problems 23-28 compute the area of the region enclosed by the graphs « INTEGRALS. e PROBLEM 23: y = 32,y = z,andz@ = 2 ¢ PROBLEM 24: y = 2z,y = =z, andy=4 ¢ PROBLEM 25: 2 = y* anda = y+ 2 e PROBLEM 26: y= e*,y=e *,andr=2 e PROBLEM 27: y= Inz,y=1-— 2, andy=2 « PROBLEM 28: y = 2°,y=2 + 6,y = 22 — 6, andy=0 ¢ PROBLEM 29: y = 2? — 22 + 2andy = 22 + 2 ARC LEnth Compute the area of the region enclosed by the graphs of the given equations. Use vertical cross-sections on Problems 1-16. ¢ PROBLEM 1 : Find the length of the graph for y = 32 + 2 on the closed interval -1< a < 2. Click HERE to see a detailed solution to problem 1. + PROBLEM? : Find the length of the graph for y = 2¥/? on the closed interval 0 <2 <4 Click HERE to see a detailed solution to problem 2 © PROBLEM 3 : Find the length of the graph for y = (3/2)z?/* on the closed interval < 2 <1. Click HERE to see a detailed solution to problem 3. * PROBLEM 4 : Find the length of the graph for y = 3 + (4/5)2°/* on the closed interval 0 < 2 <1. Click HERE to see a detailed solution to problem 4. A « PROBLEM 5 : Find the length of the graph for y = = + ay onthe closed interval 2